Let $f:N \to Y$ be a function defined as $f\left( x \right) = 4x + 3$ where $Y = \left\{ {y \in N:y = 4x + 3\;{\rm{for\;some}}\;x \in N} \right\}$ Show that f is invertible and its inverse is

$g\left( y \right) = \frac{{3y + 4}}{3}$

$g\left( y \right) = 4 + \frac{{y + 3}}{4}$

Let $f\left( x \right) = \frac{{\sqrt {\sin x} }}{{1 + \sqrt[3]{{\sin x}}}}.$ If $D$ is the domain of f, then $D$ contains

$\left( { - 2\;\pi , - \pi } \right)$

$\left( {3\;\pi ,4\;\pi } \right)$

$\left( {4\;\pi ,6\;\pi } \right)$

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 271-300 of 353 questions

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271 Medium

271 of 353

Application +4 | -1

Let

f: R \rightarrow R

be defined by

f(x)=2 x+|x|

, then

f(2 x)+f(-x)-f(x)=

View answer and explanation

Correct option: B

2|x|

Option B (

2|x|

) is correct because substituting the given function

f(x)=2 x+|x|

into the expression

f(2 x)+f(-x)-f(x)

and simplifying results in

2|x|

272 Medium

272 of 353

Understanding +4 | -1

The function

f(x)=\log \left(x+\sqrt{x^{2}+1}\right)

, is

View answer and explanation

Correct option: C

an odd function

Option C is correct because f(-x) = -f(x), which satisfies the condition for an odd function.

273 Medium

273 of 353

Application +4 | -1

f

and

g

are the function such that

(f g)(x)=(g f)(x)

for all

x

then

f

and

x

and

g

may be defined as

View answer and explanation

Correct option: D

\mathrm{f}(\mathrm{x})=x^m, g(x)=x^2+1

Option D is correct because the composition of functions

f(x) = x^m

and

g(x) = x^2 + 1

results in

(fg)(x) = (gf)(x)

for all x.

274 Medium

274 of 353

Application +4 | -1

f(x)

is defined on

\left[ {0,\;1} \right],

then the domain of

f\left( {3{x^2}} \right),

View answer and explanation

Correct option: B

\left[ { - 1/\sqrt 3 ,1/\sqrt 3 } \right]

Option B is correct because the domain of f(3x²) is found by ensuring 0 <= 3x² <= 1, which leads to the interval [-1/sqrt(3), 1/sqrt(3)].

275 Medium

275 of 353

Analyzing +4 | -1

The range of the function

f(x)=x(\log x)^2

View answer and explanation

Correct option: B

\left[0,4 e^{-2}\right]

Option B ([0,4e^-2]) is correct because the range of the function is between 0 and 4e^-2, inclusive.

276 Medium

276 of 353

Application +4 | -1

f\left( {2x + 3} \right) = \sin x + {2^x},

then

f\left( {4m - 2n + 3} \right)

is equal to

View answer and explanation

Correct option: D

\sin \left( {2m - n} \right) + {2^{2m - n}}

Option D (

\sin \left( {2m - n} \right) + {2^{2m - n}}

) is correct because it is obtained by substituting x = 2m - n into the given function.

277 Medium

277 of 353

Application +4 | -1

The solution set of the equation

[x]^2+[x+1]-3=0

, where

[\cdot]

represents greatest integer function is:

View answer and explanation

Correct option: B

[-2,-1) \cup[1,2)

Option B ([-2,-1) ∪[1,2)) is correct because it represents the solution set obtained by solving the equation with the greatest integer function.

278 Medium

278 of 353

Understanding +4 | -1

f\left( x \right) = \frac{1}{{\sqrt {\left| x \right| - x} }}

then, domain of

f\left( x \right)

View answer and explanation

Correct option: A

\left( { - \infty ,\;0} \right)

Option A (

\left( { - \infty ,\;0} \right)

) is correct because the function is defined only when

|x| > x

, which occurs when x is negative.

279 Medium

279 of 353

Understanding +4 | -1

x,y \in R,

then

\frac{1}{2}\left( {x + y + \left| {x - y} \right|} \right) = x

holds iff

View answer and explanation

Correct option: D

x \ge y

Option D (

x \ge y

) is correct because the equation holds true only when x is greater than or equal to y, ensuring the absolute value simplifies correctly.

280 Medium

280 of 353

Application +4 | -1

Let

f:N \to Y

be a function defined as

f\left( x \right) = 4x + 3

where

Y = \left\{ {y \in N:y = 4x + 3\;\mathrm{for\;some}\;x \in N} \right\}

Show that f is invertible and its inverse is

View answer and explanation

Correct option: A

g\left( y \right) = \frac{{y - 3}}{4}

Option A (

g(y) = \frac{{y - 3}}{4}

) is correct because it is the inverse function obtained by solving for x in the equation y = 4x + 3.

281 Medium

281 of 353

Application +4 | -1

The period of

f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2

is:

View answer and explanation

Correct option: C

2 n(n-1)

The correct answer is C (2n(n-1)) because it is the least common multiple of the periods of the sine and cosine functions.

282 Medium

282 of 353

Analyzing +4 | -1

f(x)=x^2+2 b x+2 c^2

and

g(x)=-x^2-2 c x+b^2

are such that

\min f(x)>\max g(x)

, then relation between

{b}

and

{c}

, is:

View answer and explanation

Correct option: D

|c| > \sqrt{2}|b|

The correct answer is

|c| > \sqrt{2}|b|

because the condition

\min f(x) > \max g(x)

leads to the inequality

c^2 > 2b^2

, which simplifies to the given relation.

283 Medium

283 of 353

Analyzing +4 | -1

The number of solutions of the equation

{\log _{x - 3}}\left( {{x^3} - 3{x^2} - 4x + 8} \right) = 3,

View answer and explanation

Correct option: A

Option A (1) is correct because only one of the potential solutions satisfies the logarithmic equation's domain restrictions.

284 Medium

284 of 353

Application +4 | -1

f(x)+f\left(\frac{1}{1-x}\right)=\frac{2(1-2 x)}{x(1-x)}

, where,

x

is a real number

x \neq 0, x \neq 1

, then the value of

f(2)

must be

View answer and explanation

Correct option: A

Option A (3) is correct because by substituting and manipulating the given functional equation, we can solve for f(2).

285 Medium

285 of 353

Understanding +4 | -1

Let

f\left( x \right) = \frac{{\sqrt {\sin x} }}{{1 + \sqrt[3]{{\sin x}}}}.

D

is the domain of f, then

D

contains

View answer and explanation

Correct option: A

\left( {0,\;\pi } \right)

Option A (

\left( {0,\;\pi } \right)

) is correct because

\sin x \ge 0

in the interval

\left( {0,\;\pi } \right)

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286 Medium

286 of 353

Application +4 | -1

Let

f(x)=a^x(a>0)

be written as

f(x)=g(x)+h(x)

, where

g(x)

is an even function and

h(x)

is an odd function then the value of the

g(x+y)+g(x-y)

is :

View answer and explanation

Correct option: A

2 g(x) g(y)

Option A (

2 g(x) g(y)

) is correct because it is the result of correctly simplifying the expression

g(x+y) + g(x-y)

using the definitions of

g(x)

and

h(x)

287 Medium

287 of 353

Application +4 | -1

Let

f\left( {x + \frac{1}{x}} \right) = {x^2} + \frac{1}{{{x^2}}},\left( {x \ne 0} \right)

then

f\left( x \right)

equals

View answer and explanation

Correct option: B

{x^2} - 2\;for\;\mathrm{ }all\;\left| x \right| \ge 2

Option B (

{x^2} - 2\;for\;\mathrm{ }all\;\left| x \right| \ge 2

) is correct because it correctly expresses f(x) in terms of x, considering the domain restriction.

288 Medium

288 of 353

Understanding +4 | -1

Let

f(x)=x, g(x)=\frac{1}{x}

and

h(x)=f(x) \cdot g(x)

, then

h(x)=1

View answer and explanation

Correct option: D

\forall x \in R, x \neq 0

Option D (

\forall x \in R, x \neq 0

) is correct because

h(x) = f(x) \cdot g(x) = x \cdot \frac{{1}}{{x}} = 1

only when

x

is a real number not equal to 0, as

g(x)

is undefined at

x=0

289 Medium

289 of 353

Understanding +4 | -1

Assertion : The function defined by

f(x)=\cos \left(x^2\right)

is a continuous function.

Reason : The cosine function is continuous in its domain i.e.,

x \in R

View answer and explanation

Correct option: B

Assertion is correct, reason is correct; reason is not a correct explanation for assertion

Option B is correct because both the assertion and reason are correct, but the reason doesn't directly explain the assertion.

290 Medium

290 of 353

Analyzing +4 | -1

f(x)=x^3+3 x^2+4 x+b \sin x+c \cos x

is one-one

\forall x \in R

, then the greatest value of

b^2+c^2

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the function being one-to-one implies its derivative is non-negative, leading to a constraint on the values of b and c.

291 Medium

291 of 353

Application +4 | -1

Suppose

f: \mathrm{R} \rightarrow(0, \infty)

be a differentiable function such that

5 f(x+y)=f(x) \cdot f(y), \forall x, y \in \mathrm{R}

. If

f(3)=320

, then

\sum\limits_{{n=0}}^5 f(n)

is equal to

View answer and explanation

Correct option: A

6825

Option A (6825) is correct because it is the sum of the function values from n=0 to n=5, given the functional equation and f(3)=320.

292 Medium

292 of 353

Application +4 | -1

The solution set of the inequation

{\log _{1/3}}\left( {{x^2} + x + 1} \right) + 1 > 0,

View answer and explanation

Correct option: C

\left( { - 2,\;1} \right)

Option C

\left( { - 2,\;1} \right)

is correct because it is the solution set of the given logarithmic inequality.

293 Medium

293 of 353

Application +4 | -1

The set of all

x

satisfying the inequality

\frac{{4x - 1}}{{3x + 1}} \ge 1

View answer and explanation

Correct option: C

\left( { - \infty ,\; - \frac{1}{3}} \right) \cup \left[ {2,\;\infty } \right)

Option C (

\left( { - \infty ,\; - \frac{1}{3}} \right) \cup \left[ {2,\;\infty } \right)

) is correct because it represents the set of all x that satisfy the given inequality after simplification.

294 Medium

294 of 353

Analyzing +4 | -1

The range of the function

f(x)=\frac{\sin \left(\pi\left[x^2+1\right]\right)}{x^4+1}

where, [] is greatest integer function, is

View answer and explanation

Correct option: C

\{0\}

Option C (

\{0\}

) is correct because the numerator

\sin(\pi[x^2+1])

is always 0, making the entire function equal to 0.

295 Medium

295 of 353

Analyzing +4 | -1

Let

f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}

be a function from

R

into

R

. The range of

f

View answer and explanation

Correct option: C

[0,1)

Option C (

[0,1)

) is the correct answer because the range of the function

f(x)=\frac{{x^2}}{{1+x^2}}

is all values greater than or equal to 0 and strictly less than 1.

296 Medium

296 of 353

Analyzing +4 | -1

f\left( x \right) = \frac{{{{\sin }^4}x + {{\cos }^2}x}}{{{{\sin }^2}x\; + {{\cos }^4}x}}

for

x \in R,

then

f\left( {2010} \right) =

View answer and explanation

Correct option: A

Option A (1) is correct because the function simplifies to 1 for all real values of x due to trigonometric identities.

297 Medium

297 of 353

Application +4 | -1

f:R \to R\;and\;g\;:R \to R

are defined by

f\left( x \right) = 2x + 3\;and\;g\left( x \right) = {x^2} + 7,

then the values of x such that

g\left( {f\left( x \right)} \right) = 8

are

View answer and explanation

Correct option: C

- 1,\; - 2

Option C (-1, -2) is correct because these are the solutions to the equation g(f(x)) = 8, found by substituting and solving the quadratic.

298 Medium

298 of 353

Application +4 | -1

The range of function

f

defined as

f(x)={ }^{6-x} P_{x-2}

View answer and explanation

Correct option: B

\{1,2,3\}

Option B (

\{1,2,3\}

) is correct because the function

{ }^{6-x} P_{x-2}

is defined only for

x=2, 3, 4

, and the corresponding values are 1, 2, and 3 respectively.

299 Medium

299 of 353

Application +4 | -1

The function

f\left( x \right) = \log \left( {\frac{{1 + x}}{{1 - x}}} \right)

satisfies the equation

View answer and explanation

Correct option: C

f\left( x \right) + f\left( y \right) = f\left( {\frac{{x + y}}{{1 + xy}}} \right)

Option C (

f(x) + f(y) = f(\frac{{x + y}}{{1 + xy}})

) is correct because it satisfies the given functional equation.

300 Medium

300 of 353

Analyzing +4 | -1

If Q denotes the set of all rational numbers and

f\left( {\frac{p}{q}} \right) = \sqrt {{p^2} - {q^2}}

for any

\frac{p}{q} \in {\cal Q}

, then observe the following statements.

I.

f\left( {\frac{p}{q}} \right)

is real for each

\frac{p}{q} \in {\cal Q}

.

II.

f\left( {\frac{p}{q}} \right)

is a complex number for each

\frac{p}{q} \in {\cal Q}

.

Which of the following is correct?

View answer and explanation

Correct option: C

I is false, II is true

The correct answer is C (I is false, II is true) because if p < q, then the square root of a negative number results in a complex number, making statement I false and statement II true.

Revision notes and key takeaways

Review FUNCTIONS in your textbook or class notes, then use this set to check your understanding.

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